Theory of Numbers (V63.0248)
Professor M. Hausner
Answers to Homework 6.
*1. Find a primitive root mod 11, Show all its powers in Z∗11 . Using your calculation, list all
primitive roots in Z∗11 .
Answer: We see that 25 = −1, so 2 is a primitive root. The powers are as follows:
n
2n
1 2
2 4
3 4 5 6
8 5 10 9
7 8
7 3
9 10
6 1
The theory says the primitive roots are 2k where (k, 10) = 1, and 1 ≤ k ≤ 10. These are
k = 1, 3, 7, 9. So the primitive roots are 2, 8, 7, and 6.
*2. Prove: The equation x11 ≡ 13 mod 175 has a unique solution mod 175. (Hint: Consider
all powers a11 where (a, 175) = 1. Use Euler’s Theorem.)
Answer: We have φ(175) = φ(52 7) = 5·4·6 = 120. So x120 = 1 in Z∗175 provided (x, 175) = 1.
We now show that if a and b are relatively prime to 175, and a11 = b11 in Z∗175 then a = b.
For in this case, we have (ab−1 )11 = 1 as well as (ab−1 )120 = 1 Therefore, (ab−1 )11s+120t = 1
for any integers s and t. Now chose s and t so that 11s + 120t = 1 Then we have (ab−1 ) = 1
or a = b in Z∗175
3. P 106/9, 10, *12, 13, *15, 18.
12. The proof is the same as in problem 2 above. Since (n, p − 1) = 1, there are integers s
and t with ns + (p − 1)t = 1. Suppose b and c are relatively prime to p, and bn ≡ cn mod p.
Then (bc−1 )n ≡ 1 mod p. But by Fermat’s theorem, (bc−1 )p−1 ≡ 1 mod p. Putting these
together, we get bc−1 = (bc−1 )ns+(p−1)t ≡ 1 mod p. Therefore b ≡ c mod p. This shows that
the n-th powers of numbers relatively prime to p are all distinct mod p, and so every element
a with (a, p − 1) = 1 is an n-th power by the pigeonhole principle.
15. We know that ah ≡ 1 mod p. Since h is the order of a, we also have ah/2 6≡ 1 mod p.
Therefore a is a zero of xh − 1. But xh − 1 = (xh/2 − 1)(xh/2 + 1) in Z∗p . Since a is not a zero
of the first factor, and is a zero of the product, it must be a zero of the second factor. Thus
ah/2 + 1 = 0 in Z∗p . This is the result.